The Central Idea
Let
\[ T \]denote the lifetime of a component or system.
The central reliability quantity is the probability that the lifetime exceeds a specified time \(t\):
\[ \boxed{ R(t)=P(T>t) }. \]The function \(R(t)\) is called the reliability function or survival function.
Lifetime as a Random Variable
A lifetime variable satisfies
\[ T\ge0. \]Different realizations of the same component type may fail at different times, so lifetime is modeled probabilistically rather than as one fixed number.
The Failure Distribution
The cumulative distribution function of \(T\) is
\[ F(t)=P(T\le t). \]This represents the probability that failure has occurred by time \(t\).
Reliability and failure probability are complements:
\[ \boxed{ R(t)=1-F(t) }. \]Interpreting Reliability
If
\[ R(1000)=0.97, \]then a randomly selected component has probability \(0.97\) of operating beyond \(1000\) time units under the assumptions of the model.
Reliability Is Time-Dependent
Reliability is not usually a single constant.
As time increases, the probability of surviving beyond that time cannot increase:
\[ t_1Initial Reliability
For a component that begins functioning at time \(0\),
\[ R(0)=P(T>0). \]In a standard continuous lifetime model,
\[ \boxed{R(0)=1}. \]Failure Density
If the lifetime distribution is continuous and differentiable, its density is
\[ f(t)=F'(t). \]Since
\[ R(t)=1-F(t), \]we also have
\[ \boxed{ f(t)=-R'(t) }. \]Probability of Failure in an Interval
The probability that failure occurs between times \(a\) and \(b\) is
\[ P(aThe Hazard Rate
The hazard rate or failure rate describes the instantaneous tendency to fail at time \(t\), conditional on having survived to time \(t\).
For a continuous lifetime distribution,
\[ \boxed{ h(t)=\frac{f(t)}{R(t)} }. \]Conditional Meaning of the Hazard
Informally, for a small interval of length \(\Delta t\),
\[ P(tThe hazard therefore measures failure intensity among components that are still operating.
Hazard Is Not a Probability
The value
\[ h(t) \]is a rate, not a probability.
It can exceed \(1\) when time is measured in sufficiently large units.
Probabilities arise after integrating hazard over time or multiplying by a very small interval in a local approximation.
Cumulative Hazard
Define the cumulative hazard
\[ H(t)=\int_0^t h(u)\,du. \]Then
\[ \boxed{ R(t)=e^{-H(t)} }. \]This provides a direct connection between the failure-rate viewpoint and the survival-probability viewpoint.
Recovering Hazard from Reliability
When \(R(t)>0\),
\[ \log R(t)=-H(t). \]Differentiating gives
\[ \boxed{ h(t) = -\frac{d}{dt}\log R(t) }. \]Constant Hazard and Exponential Lifetimes
Suppose the hazard rate is constant:
\[ h(t)=\lambda. \]Then
\[ H(t)=\lambda t \]and therefore
\[ \boxed{ R(t)=e^{-\lambda t} }. \]This is the survival function of an exponential lifetime.
Exponential Lifetime Distribution
If
\[ T\sim\operatorname{Exponential}(\lambda), \]then
\[ F(t)=1-e^{-\lambda t}, \] \[ R(t)=e^{-\lambda t}, \]and
\[ h(t)=\lambda. \]Mean Lifetime
For an exponential lifetime,
\[ \boxed{ E[T]=\frac{1}{\lambda} }. \]Thus, a lower constant failure rate corresponds to a longer mean lifetime.
Memorylessness
Exponential lifetimes satisfy
\[ \boxed{ P(T>s+t\mid T>s) = P(T>t) }. \]Conditional on still functioning at time \(s\), the remaining lifetime has the same distribution as the original lifetime.
When Memorylessness Is Reasonable
A constant hazard can be useful when failure risk is approximately stable over the relevant operating period.
It is less appropriate for systems with strong early-life defects, wear-out, fatigue, or aging effects.
Increasing Hazard
If
\[ h(t) \]increases with time, surviving components become more failure-prone as they age.
This can represent wear-out, degradation, corrosion, fatigue, or accumulating damage.
Decreasing Hazard
If
\[ h(t) \]decreases with time, early failures are more common than later failures among surviving components.
This can model infant mortality or early manufacturing defects.
The Bathtub-Curve Idea
Some engineered systems display a qualitative hazard pattern with:
- high early failure risk;
- a middle period of roughly stable risk;
- increasing late-life risk.
This is often called a bathtub curve.
Mean Time to Failure
For a nonrepairable component, the mean time to failure, or MTTF, is
\[ \boxed{ \operatorname{MTTF}=E[T] }. \]For a nonnegative lifetime,
\[ \boxed{ E[T]=\int_0^\infty R(t)\,dt } \]whenever the integral is finite.
Median Lifetime
The median lifetime \(m\) satisfies
\[ P(T\le m)=\frac12. \]Equivalently,
\[ \boxed{ R(m)=\frac12 }. \]The median and mean need not be equal.
Example: Exponential Reliability
Suppose a component has constant failure rate
\[ \lambda=0.002 \]per hour.
Its reliability after \(100\) hours is
\[ R(100) = e^{-0.002(100)} = \boxed{e^{-0.2}}. \]Example: Probability of Failure by a Deadline
Using the same component,
\[ F(100) = 1-R(100). \]Therefore,
\[ \boxed{ P(T\le100) = 1-e^{-0.2} }. \]Reliability of Independent Series Components
A series system functions only if every required component functions.
For independent components with reliabilities
\[ R_1(t),R_2(t),\ldots,R_n(t), \]the system reliability is
\[ \boxed{ R_{\text{series}}(t) = \prod_{i=1}^{n}R_i(t) }. \]Why Series Reliability Multiplies
The series system survives to time \(t\) only when all component-survival events occur.
Under independence,
\[ P(\text{all survive}) = \prod_iP(\text{component }i\text{ survives}). \]Identical Series Components
If \(n\) independent components each have reliability
\[ R(t), \]then
\[ \boxed{ R_{\text{series}}(t) = [R(t)]^n }. \]Adding required series components generally lowers system reliability.
Reliability of Independent Parallel Components
A simple parallel system functions as long as at least one component functions.
The system fails only if every component fails.
For independent components,
\[ \boxed{ R_{\text{parallel}}(t) = 1- \prod_{i=1}^{n} [1-R_i(t)] }. \]Identical Parallel Components
If \(n\) independent components each have reliability \(R(t)\),
\[ \boxed{ R_{\text{parallel}}(t) = 1-[1-R(t)]^n }. \]Redundancy can therefore increase reliability substantially.
Example: Two Components in Series
Suppose two independent components have reliabilities
\[ R_1=0.95 \]and
\[ R_2=0.90 \]at a specified mission time.
The series reliability is
\[ 0.95(0.90) = \boxed{0.855}. \]Example: Two Components in Parallel
Using the same reliabilities,
\[ R_{\text{parallel}} = 1-(1-0.95)(1-0.90). \]Therefore,
\[ R_{\text{parallel}} = 1-(0.05)(0.10) = \boxed{0.995}. \]Reliability Block Diagrams
A reliability block diagram represents logical dependence among system components.
- Series blocks represent required components.
- Parallel blocks represent redundancy.
- Mixed structures can be simplified stage by stage when independence and structure permit.
Independence Is a Modeling Assumption
Series and parallel product formulas require independence.
Shared power supplies, environmental shocks, design defects, common software, or physical proximity can produce dependent failures.
Common-cause failures can make naive redundancy calculations too optimistic.
Repairable Systems
A repairable system may alternate between functioning and failed states.
A simple state space is
\[ S=\{U,D\}, \]where \(U\) means up and \(D\) means down.
Failure and Repair Rates
Suppose an operating system fails at rate
\[ \lambda \]and a failed system is repaired at rate
\[ \mu. \]The system then forms a two-state continuous-time Markov process.
Two-State Reliability Process
The transitions are
\[ U\xrightarrow{\lambda}D \]and
\[ D\xrightarrow{\mu}U. \]Exponential up-times and repair times arise naturally from the constant-rate assumptions.
Availability
Availability is the probability that a repairable system is operational at a specified time.
Long-run availability is the long-run proportion of time the system is up.
Steady-State Availability
For the two-state failure-repair model,
\[ \boxed{ A = \frac{\mu}{\lambda+\mu} }. \]The long-run unavailability is
\[ \boxed{ 1-A = \frac{\lambda}{\lambda+\mu} }. \]Availability in Mean-Time Form
If
\[ \operatorname{MTTF}=\frac1\lambda \]and
\[ \operatorname{MTTR}=\frac1\mu, \]where MTTR is mean time to repair, then
\[ \boxed{ A = \frac{\operatorname{MTTF}} {\operatorname{MTTF}+\operatorname{MTTR}} }. \]Reliability and Availability Are Different
Reliability asks whether a system has continued functioning without failure over a specified interval.
Availability asks whether the system is operational at a specified time, allowing for the possibility that failures were repaired.
A repairable system can have moderate reliability over a long interval but high availability if repairs are rapid.
Example: Availability
Suppose
\[ \lambda=0.01 \]failures per hour and
\[ \mu=0.20 \]repairs per hour.
Then
\[ A = \frac{0.20}{0.01+0.20} = \boxed{\frac{20}{21}}. \]Mean Time Between Failures
For repairable systems, the phrase mean time between failures, or MTBF, is often used for the expected operating cycle between successive failures.
Terminology varies by engineering context, so the exact definition should be stated when calculations depend on it.
Degraded States
Real systems may have more than simply up and down states.
A state space might be
\[ S=\{\text{Full},\text{Degraded},\text{Failed}\}. \]Transition rates can then model progressive deterioration, repair, and restoration.
Redundant Repairable Systems
A system with several components can use states that record how many components remain operational.
For two identical components, one possible state space is
\[ S=\{2,1,0\}, \]representing two, one, or zero functioning units.
State-Based Reliability Modeling
Reliability can therefore be studied in two complementary ways:
- through lifetime distributions and survival functions;
- through stochastic state transitions among operating and failed conditions.
The appropriate representation depends on whether repair, degradation, and multiple component states matter.
Failure Counts
In some applications, the number of failures over time is itself modeled as a counting process.
Under constant independent failure intensity, a Poisson process may serve as an approximation for repeated failure events.
Repair and aging can make the count process more complicated.
Preventive Maintenance
Maintenance can alter reliability by changing the effective age or condition of a system.
Models may distinguish:
- minimal repair;
- perfect repair;
- imperfect repair;
- scheduled replacement.
Each assumption changes the future failure process.
Mission Reliability
In many applications, the key quantity is the probability that a system completes a mission of duration \(t\) without failure:
\[ \boxed{ P(T>t)=R(t) }. \]Mission reliability is especially important when repair during operation is impossible or undesirable.
Conditional Reliability
Suppose a component has already survived to time \(s\).
The probability that it survives an additional \(t\) units is
\[ \boxed{ P(T>s+t\mid T>s) = \frac{R(s+t)}{R(s)} }. \]For exponential lifetimes, this reduces to \(R(t)\) because of memorylessness.
Residual Lifetime
Conditional on survival to age \(s\), the remaining lifetime is
\[ T-s\mid T>s. \]Its distribution can depend strongly on \(s\) when the hazard rate changes with age.
Reliability and Expected Cost
Reliability models often feed into operational decisions involving:
- replacement cost;
- downtime cost;
- maintenance cost;
- safety penalties;
- spare-part inventory;
- warranty exposure.
Probability models provide inputs to these economic decisions but do not by themselves determine the optimal policy.
Simulation of Reliability Processes
Reliability models can be simulated by generating random lifetimes and repair times.
- Choose the initial system state.
- Generate a time to the next failure or repair event.
- Advance the simulated clock.
- Update the system state.
- Continue until the mission horizon or simulation horizon is reached.
- Record failures, downtime, and operating time.
What Simulation Can Estimate
Repeated reliability simulations can estimate:
- mission reliability;
- availability;
- downtime distributions;
- failure counts;
- repair workloads;
- probability of system-level failure;
- effects of redundancy;
- maintenance-policy performance.
Model Assumption: Lifetime Distribution
Choosing exponential, Weibull, lognormal, or another lifetime distribution is a substantive modeling decision.
The shape of the hazard rate should be consistent with the failure mechanism being modeled.
Model Assumption: Independent Components
Component reliabilities cannot automatically be multiplied unless failures are modeled as independent.
Shared environmental and structural causes can create dependence.
Model Assumption: Repair Behavior
Repair may restore a system to:
- as-good-as-new condition;
- the same effective age as before repair;
- an intermediate condition.
These assumptions materially change the future process.
Model Assumption: Operating Environment
Reliability can depend on load, temperature, vibration, maintenance quality, usage intensity, and other environmental variables.
A model with one fixed lifetime distribution may be inadequate when operating conditions vary substantially.
Common Error: Confusing Reliability with Failure Probability
Reliability is
\[ R(t)=P(T>t), \]while cumulative failure probability is
\[ F(t)=P(T\le t). \]They satisfy
\[ \boxed{R(t)+F(t)=1}. \]Common Error: Treating Hazard as a Probability
The hazard \(h(t)\) is a rate conditional on survival to time \(t\).
It is not the same as
\[ P(T\le t) \]or
\[ P(T>t). \]Common Error: Assuming Constant Failure Rate Automatically
Exponential lifetimes require a constant hazard.
Aging, wear-out, and early defects can make the hazard strongly time-dependent.
Common Error: Adding Series Reliabilities
For independent series components, system reliability is the product
\[ \prod_iR_i(t), \]not the sum of component reliabilities.
Common Error: Multiplying Parallel Reliabilities
A parallel system succeeds if at least one path succeeds.
Under independence, compute the probability that all paths fail and subtract from \(1\):
\[ R_{\text{parallel}} = 1-\prod_i[1-R_i]. \]Common Error: Confusing Reliability and Availability
Reliability usually concerns uninterrupted survival through an interval.
Availability can count a repaired system as operational again.
Repairable systems therefore require careful distinction between these two measures.
A Reliable Procedure for Lifetime Problems
- Identify the lifetime variable \(T\).
- Determine the lifetime distribution or reliability function.
- Use \(R(t)=P(T>t)\).
- Use \(F(t)=1-R(t)\) for failure by time \(t\).
- Use \(f(t)=-R'(t)\) when a density is needed.
- Use \(h(t)=f(t)/R(t)\) for conditional failure intensity.
- Check that probabilities lie between \(0\) and \(1\).
A Reliable Procedure for Series and Parallel Systems
- Draw or identify the system structure.
- Determine whether components are required in series or provide redundancy in parallel.
- Verify whether independence is a defensible assumption.
- Multiply reliabilities for independent series blocks.
- Multiply failure probabilities and subtract from \(1\) for independent parallel blocks.
- Simplify mixed systems one block at a time.
A Reliable Procedure for Repairable Availability
- Identify the up and down states.
- Identify the failure rate \(\lambda\).
- Identify the repair rate \(\mu\).
- Check that the rates use the same time unit.
- Use \(A=\mu/(\lambda+\mu)\) for the basic two-state model.
- Interpret availability as long-run operating proportion, not uninterrupted survival.
Example: Full Reliability Workflow
Suppose a nonrepairable component has exponential lifetime with
\[ \lambda=0.005 \]failures per hour.
Its reliability at \(100\) hours is
\[ R(100) = e^{-0.005(100)} = e^{-0.5}. \]Thus,
\[ \boxed{ R(100)=e^{-0.5} }. \]The probability of failure by \(100\) hours is
\[ \boxed{ 1-e^{-0.5} }. \]The mean lifetime is
\[ E[T] = \frac{1}{0.005} = \boxed{200\text{ hours}}. \]Practice 1: Reliability from Failure Probability
If
\[ F(50)=0.12, \]find \(R(50)\).
Practice 2: Exponential Reliability
A component has exponential lifetime with
\[ \lambda=0.01 \]per hour. Find \(R(20)\).
Practice 3: Mean Lifetime
For the same exponential component, find the mean lifetime.
Practice 4: Failure in an Interval
Suppose
\[ R(100)=0.90 \]and
\[ R(200)=0.75. \]Find the probability of failure between times \(100\) and \(200\).
Practice 5: Series Reliability
Two independent required components have reliabilities
\[ 0.96 \]and
\[ 0.92. \]Find the system reliability.
Practice 6: Parallel Reliability
Two independent parallel components each have reliability
\[ 0.90. \]Find the system reliability.
Practice 7: Availability
A repairable system fails at rate
\[ \lambda=0.02 \]per hour and is repaired at rate
\[ \mu=0.50 \]per hour. Find its steady-state availability.
Practice 8: Mean Time to Repair
If
\[ \mu=0.25 \]repairs per hour, find the mean repair time.
Practice 9: Conditional Reliability
Suppose
\[ R(5)=0.90 \]and
\[ R(8)=0.72. \]Find
\[ P(T>8\mid T>5). \]Practice 10: Identify the Error
A student says that a repairable system with long-run availability \(0.99\) has a \(0.99\) probability of operating for an entire year without a single failure.
Explain the error.
Practice Answers
Practice 1
\[ R(50) = 1-F(50) = 1-0.12 = \boxed{0.88}. \]Practice 2
\[ R(20) = e^{-0.01(20)} = \boxed{e^{-0.2}}. \]Practice 3
\[ E[T] = \frac{1}{0.01} = \boxed{100\text{ hours}}. \]Practice 4
\[ P(100Practice 6
Both components fail with probability
\[ (0.10)^2=0.01. \]Therefore,
\[ R_{\text{parallel}} = 1-0.01 = \boxed{0.99}. \]Practice 7
\[ A = \frac{\mu}{\lambda+\mu} = \frac{0.50}{0.02+0.50} = \boxed{\frac{25}{26}}. \]Practice 8
\[ \operatorname{MTTR} = \frac1\mu = \frac{1}{0.25} = \boxed{4\text{ hours}}. \]Practice 9
\[ P(T>8\mid T>5) = \frac{R(8)}{R(5)} = \frac{0.72}{0.90} = \boxed{0.8}. \]Practice 10
Availability measures the probability or long-run proportion of time that the system is operational, allowing repair after failures.
Reliability over a full year asks for uninterrupted survival through that interval.
The two quantities are not interchangeable.
What to Remember
Reliability begins with a lifetime random variable \(T\):
\[ \boxed{ R(t)=P(T>t) }. \]- \(F(t)=P(T\le t)\) is the cumulative failure distribution.
- \(R(t)=1-F(t)\).
- For continuous lifetimes, \(f(t)=-R'(t)\).
- The hazard rate is \(h(t)=f(t)/R(t)\).
- The cumulative hazard satisfies \(R(t)=e^{-H(t)}\).
- A constant hazard produces an exponential lifetime.
- For \(T\sim\operatorname{Exponential}(\lambda)\), \(R(t)=e^{-\lambda t}\) and \(E[T]=1/\lambda\).
- Independent series reliabilities multiply.
- Independent parallel reliability is one minus the probability that all parallel components fail.
- Zero or one repairable state is often insufficient when degradation matters.
- A two-state repairable model can use failure rate \(\lambda\) and repair rate \(\mu\).
- Its long-run availability is \(\mu/(\lambda+\mu)\).
- Reliability and availability answer different questions.
- Common-cause failures can invalidate independence assumptions.
- Simulation can combine lifetime, repair, and system-structure models when exact analysis becomes difficult.
The central modeling principle is: reliability tracks survival through time, while repairable-state models extend that framework to failure, restoration, and long-run availability.
Where We Go Next
Reliability processes illustrate how stochastic systems evolve through uncertainty while accumulating information about past performance. A deeper probabilistic idea asks when the current value of a process is already the best conditional prediction of its future value.
The next lesson, Introductory Martingales, develops fair-process intuition, conditional expectation, martingale differences, stopping ideas, and the role of martingales in stochastic-process analysis.